Conditional Probability and Bayes' Theorem
Conditional probability updates a probability once partial information is known; Bayes' theorem reverses the conditioning. DHSE questions use two-stage experiments - urns, machines producing defectives, tests with false positives - and expect a clearly identified partition.
Core rules
- P(A|B) = P(A ∩ B)/P(B), provided P(B) > 0.
- Multiplication theorem: P(A ∩ B) = P(A)P(B|A) = P(B)P(A|B).
- Independence criterion: P(A|B) = P(A), equivalently P(A ∩ B) = P(A)P(B).
- Total probability theorem: if B₁, B₂, …, Bₙ partition S, then P(A) = ΣP(Bᵢ)P(A|Bᵢ).
- Bayes' theorem: P(Bᵢ|A) = P(Bᵢ)P(A|Bᵢ) / ΣP(Bⱼ)P(A|Bⱼ).
Worked example
Urn I holds 3 red and 2 blue balls; Urn II holds 1 red and 4 blue. An urn is chosen at random and a red ball is drawn. Then P(red) = (½)(3/5) + (½)(1/5) = 0.3 + 0.1 = 0.4. By Bayes, P(Urn I | red) = 0.3/0.4 = 0.75. The prior 0.5 rises to 0.75 because Urn I is the more red-rich source.
Traps that cost marks
Reversing the conditioning and writing P(A|B) as P(B|A); omitting the total-probability denominator; choosing branches that do not form a partition; multiplying unconditional probabilities when draws are made without replacement; failing to state the prior probabilities explicitly.